3. Consider the language L (uw | u and w are strings over 0,1 and have the same number of 1s). Find the error in the following attempted proof that L is not regular: "Proof that L is not regular using the Pumping Lemma: Assume (towards a contradiction) that L is regular. Then the Pumping Lemma applies to L. Let p be the pumping length of L. Choose s to be the string 1pOP1P0, which is in L because we can choose u = 1p0p and w = 1POP which each have p 1s. Since s is in L and has length greater than or equal to p, the Pumping Lemma guarantees that s can be divided into parts x, y, z where s xyz, lul > 0, Ixy| p, and for each i-0, zyi2E L. Since the first p letters of s are all 1 and |xy p, we know that a and y are made up of all 1s. If we let i-2, we get a string ry'z that is not in L because repeating y twice adds 1s to u but not to w, and strings in L are required to have the same number of 1s in u as in w. This is a contradiction. Therefore, the assumption is false, and L is not regular